JEE Physics 2025 Question with Solution

Given: $\ell = 9 \, \text{cm}$, $d = 4 \, \text{cm}$, $\alpha = 3.1 \times 10^{-5} \, \text{K}^{-1}$, $m = 0.1 \, \text{kg}$, $C_v = 900 \, \text{J kg}^{-1}\text{K}^{-1}$, and $Q = 8.1 \times 10^2 \, \text{J}$.

Find: Change in area $\Delta A$ of the rectangular sheet.

First, determine the temperature rise using the heat relation:

$$Q = m C_v \Delta T$$

So,

$$\Delta T = \frac{Q}{m C_v}$$

Substituting the given values,

$$\Delta T = \frac{8.1 \times 10^2}{0.1 \times 900} = \frac{8.1 \times 10^2}{90} = 9 \, \text{K}$$

Now the original area is

$$A_0 = \ell \cdot d = 9 \, \text{cm} \times 4 \, \text{cm} = 36 \, \text{cm}^2 = 36 \times 10^{-4} \, \text{m}^2$$

For a rectangular sheet, the change in area is approximated by

$$\Delta A = A_0 \cdot 2\alpha \cdot \Delta T$$

Substituting the values,

$$\Delta A = 36 \times 10^{-4} \cdot 2 \cdot 3.1 \times 10^{-5} \cdot 9$$ $$\Delta A = 36 \times 10^{-4} \cdot 6.2 \times 10^{-5} \cdot 9 = 36 \cdot 6.2 \cdot 9 \times 10^{-9}$$ $$\Delta A = 2008.8 \times 10^{-9} \, \text{m}^2 = 2.0088 \times 10^{-6} \, \text{m}^2$$

Therefore, the change in area is approximately $2.0 \times 10^{-6} \, \text{m}^2$. The correct option is A.

Given: The sheet has area $A = \ell d$ and receives heat $Q$.

Find: The thermal change in area.

The initial area is

$$A = 9 \, \text{cm} \times 4 \, \text{cm} = 36 \times 10^{-4} \, \text{m}^2$$

The temperature change is obtained from

$$Q = m C_v \Delta T$$

Hence,

$$\Delta T = \frac{8.1 \times 10^2}{0.1 \times 900} = 9 \, \text{K}$$

Using the area expansion relation for an isotropic solid sheet,

$$\Delta A = A \cdot 2\alpha \cdot \Delta T$$

Substituting,

$$\Delta A = 36 \times 10^{-4} \times 2 \times 3.1 \times 10^{-5} \times 9$$

This gives

$$\Delta A \approx 2.0 \times 10^{-6} \, \text{m}^2$$

Therefore, the correct option is A.